UVA - 839

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Before being an ubiquous communications gadget, a mobile was just a structure made of strings and wires suspending colourfull things. This kind of mobile is usually found hanging over cradles of small babies.

The figure illustrates a simple mobile. It is just a wire, suspended by a string, with an object on each side. It can also be seen as a kind of lever with the fulcrum on the point where the string ties the wire. From the lever principle we know that to balance a simple mobile the product of the weight of the objects by their distance to the fulcrum must be equal. That is W_l×D_l = W_r×D_r where D_l is the left distance, D_r is the right distance, W_l is the left weight and W_r is the right weight.

In a more complex mobile the object may be replaced by a sub-mobile, as shown in the next figure. In this case it is not so straightforward to check if the mobile is balanced so we need you to write a program that, given a description of a mobile as input, checks whether the mobile is in equilibrium or not.

Input 

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The input is composed of several lines, each containing 4 integers separated by a single space. The 4 integers represent the distances of each object to the fulcrum and their weights, in the format: W_l D_l W_r D_r

If W_l or W_r is zero then there is a sub-mobile hanging from that end and the following lines define the the sub-mobile. In this case we compute the weight of the sub-mobile as the sum of weights of all its objects, disregarding the weight of the wires and strings. If bothW_l and W_r are zero then the following lines define two sub-mobiles: first the left then the right one.

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Write `YES‘ if the mobile is in equilibrium, write `NO‘ otherwise.

Sample Input 

1

0 2 0 4
0 3 0 1
1 1 1 1
2 4 4 2
1 6 3 2

Sample Output 

YES

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 
 5 int ok;
 6 
 7 int dfs() {
 8     int wl,dl,wr,dr;
 9     cin >> wl >> dl >> wr >> dr;
10     if (wl == 0) wl = dfs();
11     if (wr == 0) wr = dfs();
12     if (wl * dl != wr * dr) ok = 0;
13     else return wl + wr; 
14 }
15 
16 int main () {
17     int T;
18     freopen("1.in","r",stdin);
19     cin >> T;
20     while (T--) {
21         ok = 1;
22         dfs();
23         cout << (ok == 0 ? "NO" : "YES") << endl;
24         if (T) cout << endl;
25     }
26 }

View Code

UVA - 839

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原文地址：http://www.cnblogs.com/xiaoshanshan/p/4122680.html